y\'\'+4y\'+4y=e^-2x/x^3
要詳解喔
2006-01-17 12:27:52 · 1 個解答 · 發問者 龍斬澐 6 in 電腦與網際網路 ➔ 軟體
1. y'' + 4y' + 4y = e-2x/x3sol: 特徵方程式( characteristic equation ):r2 + 4r + 4 = 0 → ( r + 2 )( r + 2 ) = 0 → r = - 2 , - 2 ~ 重根 yh = ( c1 + c2x )e-2x = c1e-2x + c2xe-2x ~ 齊性解( homogeneous solution ) 以參數變異法( method of variation of parameters )求特解: 令 yp = u1y1 + u2y2 = u1e-2x + u2xe-2x Q = e-2x/x3 W( y1 , y2 ) =│ e-2x xe-2x │= e-4x │ - 2e-2x e-2x - 2xe-2x │ u1 =∫( - y2Q/W )dx =∫{ [ - xe-2x‧( e-2x/x3 ) ]/e-4x }dx =∫( - 1/x2 )dx = 1/x u2 =∫( y1Q/W )dx =∫{ [ e-2x‧( e-2x/x3 ) ]/e-4x }dx =∫( 1/x2 )dx = - 1/2x yp = u1y1 + u2y2 = ( e-2x/x ) - ( e-2x/2x ) = e-2x/2x ~ 特解( particular solution ) 通解( general solution ):y = yh + yp → y = ( c1 + c2x )e-2x + ( e-2x/2x ) #
2006-01-17 15:32:46 · answer #1 · answered by 龍昊 7 · 0⤊ 0⤋